I want to solve this limit:
limn→+∞ln(1+n+n3)−3ln(n)n(1−cos(1/n2))
I have proved that limn→+∞ln(1+n+n3)−3ln(n)n=0 and limn→+∞1(1−cos(1/n2))=∞ but I have indeterminate form. How can I solve that?
Answer
A long version:
limn→+∞ln(1+n+n3)−3ln(n)n(1−cos(1/n2))=limn→+∞ln(1+n+n3)−ln(n3)n⋅1n4⋅(1−cos2(1/n2))1n4⋅(1+cos(1/n2))=limn→+∞ln(1+n+n3)−ln(n3)1n3⋅sin2(1/n2)1n4⋅(1+cos(1/n2))=2⋅limn→+∞ln(1+n+n3)−ln(n3)1n3=2⋅limn→+∞n3ln(1+n+n3n3)=2⋅limn→+∞ln(1+1+nn3)n3=2⋅limn→+∞ln(1+1+nn3)n3n+1⋅(n+1)=2⋅limn→+∞(n+1)⋅ln(1+1+nn3)n3n+1=2⋅lne⋅limn→+∞(n+1)→+∞
On the 2nd line
limn→+∞sin2(1/n2)1n4=limn→+∞sin(1/n2)1n2⋅sin(1/n2)1n2=1 from
limx→0sinxx=1
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